The \(\varphi\) method of semi operators, the first half of my research

[JmsNxn]

Forgive me for minimizing doubting or being ignorant and skeptical 

Nothing personal 

but

So you have a function of 3 variables.

If you hold one variable constant you get a function of 2 variables.

It feels like you are doing that. 

I do not see the point.

H(X,s,y) = f(s,y) + F1(s,y) X + f2(s,y) x^2 + …

H(0,s,y) = f(s,y)

How is that helping ?

regards

tommy1729

OH, tommy!

No problem, I see your confusion.

That's sort of what I am doing, but it's a bit more nuanced.

We have the function:

\[
f(s,x,y) = x [s] y\\
\]

And we want to correct it so that it satisfies goodstein's equation to give us the proper solution:

\[
x \langle s \rangle y\\
\]

I should've been clearer, but without loss of generality, we can drop \(x\) from the discussion. The proof will follow the exact same for all \(x > e\). So if you can do it for \(x=3\) let's say, then the algorithm (should) converge for \(x>e\) just as well.

It's sort of like how \(b \uparrow \uparrow z\) for \(b > \eta\) is constructed using Kneser, but it's the exact same construction as when \(b=e\). So without loss of generality we can ignore the value \(b\) (within the process of construction, so to speak). So when I focus on \(f(s,y)\) rather than \(f(s,x,y)\), I'm mostly just saying we don't really care about \(x\). Set it to any value greater than \(e\) and the process works the same. That's all I meant by that. It's not a salient variable. It kinda just hangs out there.

This might bite me in the ass when we try to prove analycity in \(x\) but I don't think so...

---

If you are referring to \(\varphi_1,\varphi_2,\varphi_3\), and reducing that to two variables, that's a bit different. Which is essentially you can always write one as an expression of the other two. So it suffices to only refer to a solution pair \(\varphi_1,\varphi_2\). That doesn't align with your functional equation though, at least as I see it. If this is more what your question is asking, could you specify it better?

---

Hope that answers your question.


I started making graphs in the complex plane. Here is a graph of \(3[1.5] y\) and \(3[1.9] y\) done over \(3 \le \Re y \le 19\) and \(-8 \le \Im(y) \le 8\):

   

   


You can see it slowly turning periodic. This form of the problem looks like:

\[
x [s] y\,\,\text{for}\,\,x>e,\,0 \le s \le 2,\,|\log(y) |> 1,\,\Re(y)>0\\
\]

Which, the repelling iteration for the modified bennet operators always converges.

Edit: just wrote the domain wrong, forgot that I assumed \(\Re(y) > 0\) in my construction.